I remember having these two questions at a test I once took 4 years ago. It bugs me that I don't remember how to solve it lol.

A bowl contains 17 white, 5 red and 12 blue marbles. You blindly pick two marbles in sequence.
1) What are the odds to pick a red marble as second pick?
2) You pick up a third marble. Describe the odds for it to be a blue marble at this point.

This post was edited by Forg0tten on Dec 22 2016 05:55pm

1) First consider the first marble picked up :

- if it's white (17 chances out of 34, ie : 1 out of 2), then 5 red remains among 33, hence your odds are 5/33 ;
- if it's red (5 chances out of 34), then 4 red remains among 33 : 4/33 ;
- if it's blue (12 chances out of 34, ie : 6 out of 17), then 5 red remains among 33.

Finally :
(1/2)*(5/33) + (5/34)*(4/33) + (6/17)*(5/33) = (17*5+5*4+12*5)/(34*33) = 165 / 1122 = 5/34

2) Same thing with a third pick :

WWB -> (17/34)*(16/33)*(12/32)
WRB -> (17/34)*(5/33)*(12/32)
WBB -> (17/34)*(12/33)*(11/32)
RWB -> (5/34)*(17/33)*(12/32)
RRB -> (5/34)*(4/33)*(12/32)
RBB -> (5/34)*(12/33)*(11/32)
BWB -> (12/34)*(17/33)*(11/32)
BRB -> (12/34)*(5/33)*(11/32)
BBB -> (12/34)*(11/33)*(10/32)

Add all together : 6/17

Very much to my own surprise, your answers are identical to the ones I came up with O.O
I appreciate your time and effort!

I tried to apply the same principle but without success on this problem.

There are 2 red and 4 blue marbles available. The 6 marbles are placed in a circle in a random order. What are the odds for both red marbles to be next to two blue marbles?
I can figure out the odds for one:
4/5 on one side and 1/5 odds for 4/5 on the other side, plus 4/5th odds for 3/4 on the other side: that's 76/100 for one red marble to be envelopped by two blue ones.
But then the next one

I'd try like that : consider there are 6 free slots on a circle, each slot will receive a marble.

Place a red marble in a slot, anywhere, it doesn't matter.

Now place the second red marble in 1 of the 5 available slots.

There are 2 slots next to the first red marble already placed. If you choose one of them, the 2 red marbles will be next one to another : they won't be envelopped by blue marbles.
There are 3 other slots that you can choose for the second red marble. If you choose one of them, the 2 red marbles will be separated by blue marbles.

Conclusion : 3 chances out of 5.

Yes, I came back to this post to have the same realization haha. Not much math to this one, just logic; but I assume it can be done the way I started in my previous post, too?

Well...

Consider a red marble, and consider a side for it : 4/5 for the marble to be blue.
It it's blue, then consider the other side : 3 blue marbles remain, hence 3/4 chance for the marble to be blue.

Finally : (4/5) * (3/4) = 3/5 chances for both sides to be filled with blue marbles.

And it's done : if 1 red marble is between 2 blue marbles, then the other red marble is automatically in the same situation.

I enjoy your explanations =)